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Have the following let $P_2(R)$ denote a vector space of the real polynomial functions of degree less than or equal to two and let $B:=[P_0,P_1,P_2]$ denote the natural ordered basis for $P_2(R)$ (so $P_i(x)=x^{i})$ . Define $f\in P_2(R)$ by $f(x)=5x^{2}-2x+3$ . Write f as a linear combination of the elements of B. Compute the coordinate vector $f_B$ of f with repect to B.

$f:= 5p_2-2p_1+3p_0$ $\rightarrow$ $f_B:=[3,-2,5]$

Define $h_1,h_2,h_3\in P_2(R)$ by $h_1(x)= 7x^{2}+3x+4$, $h_2(x)= x^{2}+2x+1$, $h_3(x)= x^{2}-1$. Define $C:=[h_1,h_2,h_3]$. Assume C is an ordered basis for $P_2(R)$, construct the change of coordinate matrix, A, which converts C-coordinate to B-coordinates. Compute $A^{-1}$.

$A= \begin{pmatrix}4 & 1 & -1\\3 & 2 & 0\\7 &1 &1\\\end{pmatrix}$ $\Rightarrow$ $A^{-1}$= $\begin{pmatrix}2/16 & -2/16 & 2/16\\-3/16 & 11/16 & -3/16\\-11/16 &3/16 &5/16\\\end{pmatrix}$

Im struggling with the following Let F: $P_2(R)$ $\rightarrow$ $P_3(R)$ be the linear transformation determined by: $$F(f)(x)=6\int_{0}^{2x-1} f(t) dt$$

Compute the dimension of the kernel of F. Determine a basis for the image of F. Define $A:=[P_0,P_1,P_2,P_3]$ and compute $M_{B}^{A}(F)$, the matrix of F with respect to the given ordered bases. Determine the rank of $M_{C}^{A}(F)$.

Many thanks in advance.

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What is giving you trouble? Verify that $F(f)$ is indeed a polynomial of degree at most $4$; to compute $M^A_B(F)$, evaluate $F$ at the basis vectors in $B$, and express those images in terms of $A$; you can use the matrix $A$ (bad notation alert! You are using $A$ for two different things: the basis of $P_3(R)$, and the matrix you found earlier) to compute the matrix relative to C instead of relative to B. –  Arturo Magidin Feb 24 '12 at 21:23
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2 Answers 2

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Note that $$6\int_0^{2x-1}(at^2+bt+c)dt=16ax^3+(12b-24a)x^2+(12a-12b+12c)x-2a+3b-6c$$ so as a linear map from $P_2(R)$ to $P_3(R)$, $F$ maps $$aP_2+bP_1+cP_0\mapsto 16aP_3+(-24a+12b)P_2+(12a-12b+12c)P_1+(-2a+3b-6c)P_0$$ which should suggest both a basis for the image and how to write the matrix $M_B^A(F)$. Given a basis for the image you can determine its dimension, and since you know the dimension of $P_2(R)$ you can apply the rank-nullity theorem to get the dimension of $\ker F$. To determine the rank of $M_C^A(F)$, recall that it is the same transformation as $M_B^A(F)$, just with respect to a different basis. What does this say about the ranks of $M_C^A(F)$ and $M_B^A(F)$?

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Think I may be getting a little ahead of myself in the syllabus, trying to get a bit ahead but am a little lost. –  user24930 Feb 25 '12 at 15:30
    
I know its been a long time since this post, just concerning the last bit, would the ranks be equal? –  user24930 Apr 1 '12 at 10:00
    
@user24930 Yes. –  Alex Becker Apr 1 '12 at 14:33
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Assuming you know how to determine the basis of the kernel of a linear transformation given by a matrix, the first problem can be solved by finding the matrix of the transformation $F$. That again is a matter of figuring out what $F$ does to basis vectors; consider a linear transformation $F: V \to W$ and let $\{u_i\}_{i = 1}^{m} \subseteq V$ be a basis for the domain $V$ and let $\{v_j\}_{j = 1}^{n} \subseteq W$ be a basis for the codomain $W$. To determine the matrix element $f_{ij}$ in the matrix $F = [f_{ij}]$, take the $i$'th basis vector $u_i$ in your chosen basis for the domain, apply the transformation $F$ and write the result in terms of the basis for the codomain. The coefficient $f_{ij}$ is simply the constant in front of the $j$'the basis vector, i.e. $$Fu_i = \sum_{j = 1}^{n} f_{ij} v_j $$  

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